Here is an example of a very classic SAT algebra problem. There are two ways to solve this problem; one is very intuitive and simple, but potentially very slow. The other is rooted in simple algebraic methodology (I promise you’ve done it a million times in school by now) and is extremely fast and precise. Let’s take a look at how basic algebra can help you save some major time on the math section.
If x + y = 80 and x < 30, which of the following must be true?
(A) x > 0
(B) y < 50
(C) y = 50
(D) y > 50
(E) y < 80
Step 1: Identify the Problem
The problem is asking us to determine which of the equalities has to be true, given the equation and the equality in the question. Only one of the answer choices is ALWAYS true for the given conditions. The word “must” is important in this question — it means that you are looking for the answer choice that HAS TO BE true, not just “can be” true.
Step 2: Gather Information
The relevant information for this problem is given in the problem. The first piece is x + y = 80, and the second piece is x < 30.
Step 3: Use the Information to Solve the Problem
As I said earlier, there are two ways to solve this problem. The simpler and more intuitive way that most people will immediately jump to is basically trial and error. If we test each of the answer choices, we can arrive at the correct answer by eliminating the ones that we discover are not always true.
Method 1: Elimination by Testing
Remember — we are looking for the answer choice that MUST be true, given the two equalities in the question. So, if we can show that an answer choice is not necessarily true, then we can eliminate it. In order to do that, we will pick numbers that don’t satisfy the answer choice equality, and then test them — if they still work with the original two conditions, then we have disproved it, and that answer choice is not correct.
For (A) the question we are asking ourselves is, “Does x HAVE to be greater than 0?” We can test this question by choosing x = -10. Does it still match the original conditions?
First, let’s check x < 30. Plug in -10 for x, and we get -10 < 30. This is true, so that part is OK.
Now, let’s plug x = -10 into the equation. After solving the equation for y, we see that y = 90. Well, there are no conditions specifying what y has to be, so y = 90 is also fine. So now we know that x does NOT have to be greater than 0 in order to satisfy the two original equalities. This means that (A) does not have to be true.
Now let’s check (B). We ask ourselves, “Does y HAVE to be less than 50?” Let’s pick the value y = 100. If we plug y = 100 into the equation, we can see that x = -20. -20 is less than 30, so all the original conditions still work. Thus, (B) is not the answer either.
For (C), let’s pick the value y = 100 again. We already know from (B) that when y = 100, x = -20, and we already know that -20 is less than 30. So (C) is also not the correct answer.
For (D), let’s pick y = 0. If we plug that in, we get x = 80. 80 is not less than 30. So far so good. Let’s pick another number. Since we picked a relatively small number last time, let’s pick the biggest number we can this time. The biggest number that still fits the condition y < 50 is 49. If we plug in y = 49, we get x = 31. Once again, 31 is not less than 30, so we are good. Finally, let’s pick a number even smaller than 0, like -10. If we plug y = -10 into the equation, we get x = 90. Looking at these three answers, we can see that the smaller y gets, the larger x gets. Thus, we can conclude that y MUST be greater than 50, otherwise the original equalities do not work. So (D) works as an answer.
Finally, let’s check (E) by picking y = 100 again. Because of answers (B) and (C), we already know that x = -20, so (E) is disproven as well.
This method works and is very intuitive for most people, but it’s a little bit slow, and it depends on the numbers that you choose. It’s also very common for students to trip up and test the wrong numbers. Obviously you don’t have time to pick too many numbers on the SAT and sit there trying to figure out whether or not you’re performing tests on the correct numbers. There must be a faster way that can definitively get you the correct answer…
Method 2: Algebraic Manipulation
With just a little bit of simple algebraic manipulation, you can very quickly arrive at the correct answer with 100% certainty. Let’s solve the first equation for x. We get
x = 80 – y
Now that we have an expression for x, we can substitute it into the equality. We get
80 – y < 30
Now if we solve this equality for y, we get
50 < y, which can be rewritten as y > 50. This is answer choice (D), and it is the only answer choice that can be definitively and mathematically proven to be true using simple algebraic manipulation.
Not all of us are mathematically minded, and not all of us would have known to solve this problem that way. It’s not complicated math, right? But somehow, some of us just don’t realize that this is the most effective way to solve this problem. If you are wondering how you will be able to recognize a similar situation on your SAT, the best answer is practice paired with instruction. Practice alone isn’t enough — if you practice something a million times the wrong way, all that practice isn’t going to help you. Instruction alone isn’t enough either — just because you read this blog post doesn’t mean you will immediately be able to recognize the scenario in the future. SAT test preparation has many critics, and some of the points they make are fair, but there are many benefits to taking a course taught by people who know the best ways to solve these problems and can help you practice solving them the correct way over and over until you understand it. This problem is an excellent example. You already have the skills needed to solve the problem quickly, we’re just here to help you recognize when you need to use those skills. And we have tons of problems at our disposal to help you practice using the best methods over and over again until it becomes second nature.
Now, this is just a blog post, so all I can do is give you some tips. Notice that the problem is basically all equations. This should tip you off to the fact that it’s an algebra problem. And if it’s an algebra problem, you can bet that it will be testing your understanding of basic algebraic principles like solving equations and substitution. When you see a problem that is obviously about algebra, think about how you might be able to manipulate the equations to get the answer you want. In this problem, you literally only had three steps:
1. Solve the first equation for x.
2. Substitute x into the second equation.
3. Solve the second equation for y.